A right triangle has a leg of length 10. The angle opposite this side has an angle θ = π/4 rad. What is the length, L, of the other leg?
If you recognize θ = π/4 rad = 45°, then you may realize this is a right isosceles triangle and the legs are the same, making the length also 10. Alternatively, you would also get this answer by recalling that tangent = opposite/adjacent. To get the length of the side, it follows that L = opposite * (adjacent/opposite) = opposite * (1 / tan θ) = 10 * (1 / tan(π/4)) = 10
A container of water weighing 50 N is resting on a scale (the scale has been tared to the container). You drop a 10 N copper weight (density ~9.0 g/cm^3) into the water, which sinks to the bottom. As you reach your hand into the water to grab the weight (but do not touch anything except the water), you notice the water level now reads 5.7 L. What does the scale read?
This problem can be broken into several smaller parts. To find the total scale reading, we must account for all forces acting on the scale. The more obvious forces are the weight of the water and the weight of the copper, but we also need to include the buoyant force exerted on your hand because that force is acting equal and opposite on the water. Like a solid object pressing back against your hand if you pushed it, the water pushes on your hand too (and your hand pushes on the water). This force can be determined by the weight of the water your hand displaces. Because the copper weight sunk and did not float, we know that the volume displaced only depends the volume of the object, not the weight. With the density ρ = 9.0 g/cm^3 = 9 kg/L, we can determine that the volume of the copper weight was V_copper = 10 N/[(9.81 m/s^2)*(9.0 kg/L)] = 0.1133 L We know that the water weighs 50 N, which assuming ρ = 1.0 g/cm^3 = 1 kg/L tells us that the initial volume was V_water = 50 N/[(9.81 m/s^2)*(1.0 kg/L)] = 5.097 L If the water level now reads 5.7 L, your hand must have displaced V = (5.7 L) - (5.097 L) - (0.1133 L) = 0.4897 L The buoyant force exerted on your hand is then equal to the weight of the water displaced F = (9.81 m/s^2)*(0.4897 L)*(1 kg/L) = 4.8 N Summing the forces, we have scale reading = (50 N) + (10 N) + (4.8 N) = 65 N
A pilot is flying a propeller aircraft and is attempting to maximize their range, but needs to maintain the current altitude. How should the pilot adjust airspeed, if at all, to continue flying at the best range condition?
The best range flight condition corresponds to a specific value of CL (coefficient of lift), where the ratio L/D (lift over drag) or CL/CD (coefficient of lift over coefficient of drag) is maximum. This represents the flight condition where the least amount of energy is expended for each unit distance traveled. As the airplane continues flying, its weight decreases as a result of the fuel used, and therefore the lift required to maintain flight also decreases since L = W in steady level flight. Since the lift produced is dependent on both CL and V (airspeed) where L = 1/2*ρ*S*CL*V^2 the airspeed must be decreased to accommodate the decreased need for lift. Remember that the density, ρ, remains constant (due to constant altitude), the surface area, S, doesn't change (wing size is fixed), and the pilot wants to maintain the same CL to maximize range. We can see then that it is necessary for the pilot to decrease airspeed as the flight continues to maximize their range. It is important to note that the airspeed corresponding to the best range condition is not the same as best cruise, where the aircraft has the greatest endurance (time-wise, not distance-wise).